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##### Mathematics (2015) QUIZ STUDY

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Simplify 1¼ ÷ (2 ÷ ¼) of 28

1 \( \frac{3}{8}\)

2 \( \frac{3}{4}\)

4 \( \frac{3}{8}\)

3 \( \frac{1}{5}\)

1¼ ÷ [ 2 ÷ ¼] of 28

Apply BODMAS rules

\( \frac{5}{4}\) ÷ [ 2 ÷ ¼ \(\times\) 28 ]

\( \frac{5}{4} ÷ \frac{2}{7} \)

\( \frac{5}{4} \times \frac{7}{2} \)

\( \frac{35}{8}\)

= \(4 \frac{3}{8}\)

Factorize x^{2} + 9x + 20

(x - 5) ^{2}

(x + 5)(x + 4)

(x + 5)(x + 3)

(x + 3) ^{2}

\( x^2 + 9x + 20 \)

Find the two numbers whose product is 20 and its sum is 9}

\( 5x \times 4x = 20x^2\)

\(5x + 4x = 9x\)

\((x^2 + 5x) + (4x + 20)\)

\(

x(x + 5)+ 4(x + 5)

= (x + 5)(x + 4)\)

If three staff of QuizzerWeb Inc. agreed to share their salary arrears in the ration of their ages, which are 18 years, 20 years, 22 years respectively. If the sum of the money collected is N120,000.00K, How much does the second staff received?

N36,000

N44,000

N40,000

N15,000

X and Y are two sets such that nX = 15, nY = 12 and n{- Y} = 7. Find -{X - Y}

21

225

15

20

n(X - Y) = nx + ny - n(xX - Y)

= 15 + 12 - 7

- n(X - Y) = 20

Find the x and y intercepts of the graph of 3x - z \(\leq\) 9

(3, -9)

(-3, 9)

(-3, -9)

(-3, 0)

Starting from 3x - z \(\leq\) 9

x = 0, substitute the value of x (i.e. x = 0) into the equation 3x - z \(\leq\) 9

3(0) - z \(\leq\) - 9

z \(\leq\) - 9

If z = 0

Then, 3x - 0 \(\leq\) 9

3x \(\leq\) 9

x \(\leq\) 3

Intercept of x and z i.e (x,z)

= (3, -9)

Simplify log_{10}1.5 + 3 log_{10}2 - log_{10}0.3

log_{10}4

log_{10}40

log_{10}-40

log_{10}4^{-}

log_{10}1.5 + 3 log_{10}2 - log_{10}0.3

log_{10}1.5 + log_{10}23 - log_{10}0.3

[log_{10}(1.5 \(\times\) 23) ÷ log_{10}0.3]

[log_{10}(15/10 \(\times\) 8 ) ÷ log_{10}( 3/10)]

log_{10}(15/10 \(\times\) 8 \(\times\) 10/3 )

log_{10}40

Find the total surface area of a cylinder of base radius 5cm and length 7cm ( π = 3.14)

17.8cm^{2}

15.8cm^{2}

75.4cm^{2}

54.7cm^{2}

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 \(\times\) 3.14(7+%=5)

2 \(\times\) 3.14 \(\times\) 12

= 75.4cm (1DP)

If x + y = 90 simplify (sinx + siny)^{2 - }2sinxsiny

1

\( 0 \)

2

-1

A man with an annual salary of N2000, has allowances of N600. If Income Tax is 5%. How much tax does he pay each year?

15

20

30

25

His annual salary = N2000

His allowances = N600

So his taxable income = Annual salary - allowance

= N2000 - N600

= N1400

He pay at 5%

Then, his allowance income tax 5/100 \(\times\) 600 = N30

Solve the equation \( 3x^2 - 4x - 5 = 0 \)

x = 1.75 or - 0.15

x = 2.12 or - 0.79

x = 1.5 or - 0.34

x = 2.35 or -1.23

The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms

nth = 3

nth = 4

nth = 5

nth = 6

Nth term of a linear sequence (AP) = a+(n - 1)d

first term = 6, last term = 10 sum - 40

i.e. a = 6, l = 10, S = 40

sn = n ÷ 2(2a + (n - 1)d or Sn = ÷2 (a + b)

Sn = n ÷ 2(a + l)

40 = n ÷ 2(6 + 10)

40 = 8n

8n = 40

8n = 40

N = 40/8

= 5

The nth term = 5

Find the equation of a line which is form origin and passes through the point (-3, -4)

y = \( \frac{3x}{4} \)

y = \( \frac{4x}{3} \)

y = \( \frac{2x}{3} \)

y = \( \frac{x}{2} \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)

=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

Make x the subject of the equation

s = 2 + \(\frac{t}{5} \)(x + -y)

x = 5[(s - 2) ÷ t] + 3yt

x = 25[(s - 2) ÷ t] - 3ty

x = [1 ÷ (s - 2)3ty]

x = [5(s - 2) 2 ÷ 3ty] \(\times\) t

2 + t/5(x + 3/5y ) = s

t/5(x + 3/5y ) = s - 2

[(t(5x + 3y) ÷ 25)] = s - 2

t(5x + 3y) = 25 (s - 2)

5x + 3y = [(25(s - 2))÷ t]

5x = [(25(s - 2))÷ t] - 3y

Divide both sides by 5

x = [25(s - 2) - 3ty)]÷ 5t

_{5}20 = x, find x

_{5}20 = x

5x = 20(Take log

_{10}of both sides)

log

_{5x}= log

_{20}

xlog

_{5}= log

_{20}

x= [log

_{20}÷ log

_{5}]

[1.30103 ÷ 0.69897]

x = 1.861

Find at which rate per annum simple interest N525 will amount to N588 in 3 years.

3%

2%

5%

4%

I = A - P

= N588 - N525

-I = N63

I = PRT ÷ 100

R = [100I ÷PT]

R = [(100 \(\times\) 63 ) ÷ (525 \(\times\) 3)]

= (6300 ÷ 1575) = 4

- The rate = 4 %

Find the range of values of t which satisfies the inequality.

2t - 1 \(\leq\) 3 and 2 - t \(\geq\) 5

-3 \(\leq\) t \(\leq\) 1

-2 \(\leq\) t \(\leq\) 3

-3 \(\leq\) t \(\leq\) 4

-3 \(\leq\) t \(\leq\) 2

2t - 1 \(\leq\) 3 and 2 - t < 5

2t \(\leq\) 3 + 1 and 2 - 5 \(\leq\) t

2t \(\leq\) 4 and - 3 \(\leq\) t

t \(\leq\) 2 and - 3 \(\leq\) t or t \(\geq\) - 3

Combining the two ,starting from the last

-3 \(\leq\) t \(\leq\) 2

Given that A = {1, 5, 7}

B = {3, 9, 12, 15}

C = {2, 4, 6, 8}

Find (A \(\cup\) B) \(\cup\) C

{1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

{1, 2, 3, 5, 6, 8, 12, 15}

{2, 4, 5, 9, 12, 15}

{1, 5, 6, 7, 8, 9, 12, 15}

{A \(\cup\) B \(\cup\) C}

{A \(\cup\) B} = {1, 3, 5, 7, 9, 12, 15}

C={2,4,6,8}

{A \(\cup\) B} \(\cup\) C = {1, 3, 5, 7, 9, 12, 15} \(\cup\) {2, 4, 6, 8}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

{A \(\cup\) B} \(\cup\) C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton's is 2cm, find the extension produced by a tension of 12 newton's.

2

1

\( 0\)

3

Factorize \( x^2 - 2x - 15 \)

(x + 3)^{2}

(x + 5)(x - 3)

(x - 5)^{2}

(x - 5)(x + 3)

x^{2} - 2x - 15

(x^{2} - 5x) + (3x - 15)

x(x - 5)+ 3(x - 5)

(x - 5)(x + 3)

If A = (-3,5) and B = (4,-1) find the co-ordinate of the mid point

2, ½

½ , 2

1, ½

0, 2

[(325 \(\times\) 5 \(\times\) 3) ÷ (100)]

N(195 ÷ 4)

N48.75K

^{2}and give your answer correct to 4 significant figures

^{2}= 0.09 \(\times\) 0.09

= 0.0081

= 0.008100 to 4 significant figures

Please, start counting first from the non-zero digits i.e. 8

= \( 1 ÷ (x^2 + 3x + 2) + [1 ÷ (x^2 +5x + 6)]\)

= \( [1 ÷ ((x^2 + x) + (2x + 2) )] + [1 ÷ ((x^2 + 3x) + (2x + 6) )] \)

= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]

= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]

=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)

Using the L.C.M

=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)

=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)

= \( \frac{2 }{(x + 1)(x + 3)} \)

One bag contain 3 blue and 5 red balls, another bag contain 2 blue and 4 red balls respectively. One ball is drawn for each bag. What is the probability both balls are blue

\( \frac{2}{15} \)

\( \frac{3}{24} \)

\( \frac{3}{21} \)

\( \frac{3}{28} \)

One bag with 3 blue and 5 red

Pr (H) = 3/8Pr (r) 5/8

Another bag with 2 blue and 4 red (note one ball is drawn from each bag)

i.e. prob.=8 - 1 = 7

prob. (2 ball both are blue=pr (1st blue and 2nd blue)

prob. 2 balls both blue = 3/8 \(\times\) 2/7

= \( \frac{3}{28} \)

Given that A = {3, 4, 1, 10, ? }

B = {4, 3 ?, ?, 7}.

Find A ∩ B

{}

{?, 1, 2}

{3, 4, ?}

{1, 3, ?}

A ∩ B = {3, 4, ?}

Find x if 132_{x} = 70 _{eight.}

5

3

6

1

132_{x} = 70_{8}

\( 1 \times x^2 + 3 \times x^1 + 2 \times x^0 \)

\( 7 \times 8^1 + 0 \times 8^0 \)

\( x^2 + 3x + 2 - 56 = 0 \)

\( x^2 + 3x - 54 = 0 \)

x(x + 9)-6(x + 9) = 0

(x + 9)(x - 6) = 0

Either (x + 9) = 0 or (x - 6) = 0

x = - 9 or +6

The positive values for x = 6

The base number for 132_{x} = 132_{6}

A man sells his new brand car for N420,000 at a gain of 15%. What did it cost him?

N410,000

N365, 217

N157, 250

N257,000

Cost Price:

Selling price = 100 : (100 + 15)

= 100 : 115

Cost Price

= 100 ÷ 115 of selling price (i.e N420,000)

= 100 ÷ 15 \(\times\) N420,000

= N365,217.4

Cost Price = N365, 217

Find the value of x if \( [ 1 ÷ 64^{(x + 2)}]= [4^{(x ? 3)} ÷ 16^x ] \)

\( \frac{3}{2} \)

\( \frac{2}{3} \)

\( \frac{1}{3} \)

\( -\frac{3}{2} \)

\( [ 1 ÷ 64^{(x + 2)}]= [4^{(x - 3)} ÷ 16^x ] \)

\( 64^{-(x + 2)} = [4^{(x - 3)}] ÷[16^x] \)

breakdown 4,16,64 into a small index no

\( 2^{-6(x + 2)} = 2^{2(x - 3)} ÷ 2^{4(x)} \)

\( 2^{-6x- 12} = 2^{2x - 4x - 6} \)

\( 2^{-6x -12} = 2^{-2x - 6} \)

- 6x - 12 = - 2x - 6

Collect the like term

-6x + 2x = -6 + 12

-4x =6

x = \( \frac{6}{4} \)

x = \( \frac{-3}{2} \)

A pipe made of metal 10cm thick has an external radius of 11cm. find the area of metal in making 2.4cm of pipe

21πcm^{2}

21πcm^{2}

15πcm^{2}

17πcm^{2}

The external radius = 11cm

The internal radius = 10cm

The area of cross section = π(π^{2} - 10^{2})

= π (11 + 10)(11 - 10)

π(21)(1)

= 21πcm^{2}

Solve x^{2} - 2x - 3 = 0

x = 2 or 1

x = 3 or -1

x = 1 or 0

x = 1 or 2

x^{2} - 2x - 3 = 0

-3x + x = - 2x

- 3x \(\times\) x = - 3x^{2}

(x^{2} - 3x)(x - 3) = 0

(x - 3)(x + 1) = 0

Either (x - 3) = 0 or (x + 1) = 0

x = 3 or - 1

The volume of a cylinder whose height is 4cm and whose radius 5cm is equal to (π = 3.14)

3.13cm^{2}

145cm^{2}

314cm^{2}

214cm^{2}

€œExplanation

V = πr^{2}h

V = ?, h = 4cm, r = 5cm

V = 3.14(5)^{2} (4)

V = 3.14(25)(4)

V = 3.14 \(\times\) 1200

= 314cm^{2}

\( \int x^2 \text{Sin}x \Delta x \)

-Cos x + 2xSin x + C

-Cos x + 2

x^{2}Cos x + 2xSin x + 2Cos x+ C

-x^{2}Cos x + 2xSinx + 2Cos x + C

Let u= x^{2}

du = 2xdx

Let v = sinx

-dv = - sin x i.e. v = cos x

- udv = uv - - vdu

- x^{2} Sin xdx = - x^{2} Cos x - - (- Cos x)2 \(\times\) dx

x^{2} Cos x + 2 - x cos x^{1} dx

= - x^{2} Cos x + 2 - x Cos x^{2} dx

2 - u Cos xdx

Let u = x

du = dx

dv = Cos x

-dv = - Cos x

- x^{2} Sin x^{1} dx = - x^{2} Cos x + 2Sin x - 2Cos x + C

- x^{2} Sin xdx

= - x^{2} Cos x + 2xSin + 2Cos x + C

The probability of an event A is 1/5. The probability of B is 1/3 . The probability both A and B is 1/15. What is the probability of either event A or B or both

\( \frac{2}{15} \)

\( \frac{3}{4} \)

\( \frac{7}{15} \)

\( \frac{1}{15} \)

Prob(A) = \( \frac{1}{5} \) , Prob(B) = \( \frac{1}{4} \), Prob (A ∩ B) = \( \frac{1}{15} \), Prob (A ∩ B) = ?

Note:

I. the probability is either event of A or B or both.

The formula using is prob (A \(\cup\) B) = -prob(A) + prob(A) \(\cup\) prob(A ∩ B)

II. But if the probability of both outcomes A and B

The formula using is Prob(A ∩ B) \(\cup\) prob(A) + prob(B) + prob (A \(\cup\) B)

In this question, A or B, Prob (A \(\cup\) B): A and B, Prob (A \(\cup\) B)

prob (A \(\cup\) B) = prob(A) + prob(B) \(\cup\) Prob (A ∩ B) is used.

prob (A \(\cup\) B) = \( \frac{1}{15} \) + \( \frac{1}{3} \) + \( \frac{1}{15} \)

= (3 + 5 + 1) ÷ 15

= \( \frac{7}{15} \)

Simplify 4 - [(390695T - 8)½)]

1/T

2T

5/T

T/5

4-(390625T - 8)½

[(390625T - 8)½]¼

[((390625)½ \(\times\) T - 8)½]¼

(- 390625 \(\times\) T - 4 )¼

(625 \(\times\) T -¼)¼

(54 \(\times\) T -4)¼

= (54)¼ \(\times\) (T - 4 )¼

= 5 \(\times\) T - 1

= 5 \(\times\) 1/T

= 5/T

_{10}= (y)

_{6}. Find x

_{6}= 123.34

_{6}

_{6}= 424.5

_{6}

_{6}= 122.43

_{6}

_{6}= 124.45

_{6}

_{10}= (y)

_{6}

\( \begin{array}{c|c}

6 & 159 \\

\hline

6 & 26 \text{ rem 3} \\

6 & 4 \text{ rem 2} \\

& 0 \text{ rem 4} \\

\end{array}\uparrow \)

159

_{10}= 423

_{6}

Then Convert 0.75 to base 6

0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5

0.5 x 6 = 3.00 423 + 1.5 =424.5

159.75 = (424.5)

_{6}

(159)

_{10}= (424.5)

_{6}

A man bought a car for N800 and sold it for N520. Find his loss per cent

15%

25%

35%

10%

Percentage Loss = (actual loss ) ÷ (Cost price) \(\times\) 100

Actual loss = Cost Price - Selling Price (sold price)

= N800 - N520 = N280

Percentage loss = (280 ÷ 800) \(\times\) 100 = 35%

⇒ His loss percent = 35%

Simplify \(6\frac{1}{12} - 2 ¾ + 1 ½ \)

\( 3 \frac{5}{6}\)

\( 4 \frac{5}{6}\)

\( 2 \frac{1}{6}\)

\( \frac{1}{3} \)

6½ - 2¾ + 1½

73/12 - 11/4 + 3/2

[(73 - 33 + 18) ÷ (12)]

58/12

29/6

\( 4\frac{5}{6}\)

Determine the third term of a geometrical progression whose first and second term are 2 and 14 respectively

1458

1485

1345

1258

1st G.P. = a =2

2nd G.P. = ar - 1 = 54

2(r) = 54

r = 54/2 = 27

r = 27

3rd term = ar^{2} = (2) (27)^{2}

2 \(\times\) 27 \(\times\) 27

= 1458

Given that S and T are sets of real numbers such that S = {x : 0 \(\leq\) x \(\leq\) 5} and T = {x : - 2 < x < 3} Find S \(\cup\) T

-3 < x \(\leq\)3

-2 < x \(\leq\) 5

2< x \(\geq\) - 5

-1 \(\leq\) x \(\geq\) 2

S = {0, 1, 2, 3, 4, 5}

T = {- 1, 0, 1, 2}

S \(\cup\) T = {- 1, 0, 1, 2, 3, 4, 5 }

- - 2 < x \(\leq\) 5

If an investor invest N450,000 in a certain organization in order to yield X as a return of N25,000. Find the return on an investment of N700,000 by Y in the same organization.

N14,950.50K

N25,150.30K

N15,000.00K

N38,888.90K

[Return ÷ Investment] as a ratio ;

i.e The Ratio is Return : Investmen

[(Return1÷ Investment1 ) = (Return2÷ Investment2)]

R1 = N 25000

R2 =?

I1 = N450,000,

I1 = N 700000

(25000 ÷ 450000) = (R2 ÷ 700000)

R2 = [(25000 \(\times\) 700000 ) ÷ 450000]

= N38,888.90K

The return on a investment of Y = N38888.90K

_{7}17

_{7}17

= [log 17 ÷ log7]

= [1.2304 ÷ 0.8451]

[10

^{0.0899}÷ 10

^{1.9270}]

= 1.455(antilog)

10011_{2} + *****_{2} + 11100_{2} + 101_{2} = 1001111_{2}

1111_{2}

11011_{2}

10111_{2}

11001_{2}

Convert the binary to base 10 and they convert back to base two

10011_{2} + xxxxx_{2} + 11100_{2} + 101_{2} = 1001111_{2}

(1 \(\times\) 2^{4} + 0 \(\times\) 2^{3} + 1 \(\times\) 2^{2} + 1 \(\times\) 2^{1} + 1 \(\times\) 2^{0}) + xxxxx_{2} +(1 \(\times\) 2^{4} + 1 \(\times\) 2^{3} + 1 \(\times\) 2^{2} + 0 \(\times\) 2^{1} + 0 \(\times\) 2^{0}) + (1 \(\times\) 2^{2} + 0 \(\times\) 2^{1} + 1 \(\times\) 2^{0})

= (16 + 0 + 0 + 2 + 1) + xxxxx_{2} + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)

=(64 + 0 + 0 + 8 + 4 + 2 + 1)

19 + xxxxx_{2} + 33 = 79

xxxxx_{2} + 52 = 79

xxxxx_{2} = 79 - 52

xxxxx_{2} = 27_{10}

\( \begin{array}{c|c}

2 & 27 \\

\hline

2 & 13 \text{ rem 1}\\

2 & 6 \text{ rem 1}\\

2 & 3 \text{ rem 0}\\

2 & 1 \text{ rem 1}\\

& 0 \text{ rem 1}\\

\end{array}\uparrow \)

27_{10} = 11011_{2}

Therefore xxxxx_{2} = 27_{10} = 11011_{2}

Evaluate log_{5}(\( y^2x^5 ÷ 125b½) \)

2 log_{5}y + 5log_{5} y^{2} - 3

log_{5} y^{2} + 5log_{5} x + 3

25log_{y} 5 + 3

2log_{5}y + 5log_{5}x - ½ log_{5}b - 3

If y = \((x^2 + 3x + 1) ÷ (3x + 4).\) Find dy/dx

\( (3x^2 + 8x + 7) ÷ (3x + 4)^2 \)

\( x^2 + 2x + (1 ÷ (x + 1)^2) \)

\( 2x^2 + 4x + (1 ÷ (2x + 4)^2) \)

\( x^2 + 3x + (2 ÷ (x + 3)^2) \)

dy/dx \( (4x^3 + 3x^2 + 2x + 1) \)

= \( 12x^2 + 6x + 2 \)

= \( 12x^2 + 6 + 2 \)

Divide both sides by 2

\( \frac{12x^2}{2} + \frac{6x}2 + \frac{2}{2} \)

= \( 6x^2 + 3x + 1 \)

dy/dx

= \( 6x^2 + 3x + 1 \)

Integral -\( (5x^3 + 7x^2 - 2x + 5)\)dx

\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)

\( \frac{5x}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)

\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)

\( \frac{2x^2}{3} + \frac{x}{5} - C \)

\(- [(x^2 + 4x^2 + 1 ) ÷ x^2]dx = - (x^3/x^2)dx + -(1/x^2)dx \)

-\( - xdx + - 4^{-1} or + - (1/x^2)dx\)

= \( x^2/2 - 4x - 1/x^2 + C \)

Find the area of the curved surface of a cone whose base radius is 3cm and whose height is 4cm (π = 3.14)

17.1cm^{2}

27.2cm^{2}

47.1cm^{2}

37.3cm^{2}

Find the slant height

\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)

\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)

\( l^2 = 25 \)

Squaring both sides

l = 5cm

The area of curved surface (s) =π(3)(5)

15π = 15 \(\times\) 3.14

= 47.1cm^{2}

Simplify \( \frac{1}{(x + 1)} + \frac{1}{(x - 1)} \)

\(\frac{2x}{(x + 1)(x-3)} \)

\(\frac{2}{(x + 1)(x-1)} \)

\(\frac{2x}{(x + 1)2}\)

2x(x+1)2

[1 ÷ (x+1)] + [1 ÷ (x - 1)]

= ((x - 1) + [(x + 1)) ÷ (x+1)(x - 1)]

Using the L.C.M.

= (x - 1 + x + 1) ÷ (x + 1)(x - 1)

= (x + 2 - 1 + 1) ÷ (x + 1)(x - 1)

= 2x ÷ (x + 1)(x - 1) =2x ÷ (x + 1)(x - 1)

The area of a circle of radius 4cm is equal to (Take π = 3.142 )

10.3cm^{2}

15.7cm^{2}

50.3cm^{2}

17.4cm^{2}

Area of a circle (A) = πr^{2}

Radius = 4cm

π = 3.142

A = 3.142 \(\times\) (4)^{2}

= 3.142 \(\times\) 16

= 50.272

A = 50.3cm^{2} (1dp)

The area of an ellipse is 132cm^{2}.The length of its major axis is 14cm.Find the length of it minor axis

10.5cm

5cm

10cm

12cm

The area of an ellipse (e) =πab/4

Let a rep. the length of its major axis = 14cm

Let b rep. the length of its minor axis = ?

π = 3.142

Area of an ellipse = 132cm^{2}

132 = (π14 \(\times\) b) ÷ 4

132 = (3.142 \(\times\) 14b) ÷ 4

3.142 \(\times\) 14b =132 \(\times\) 4

b = (1324 \(\times\) 4) ÷ (3.142 \(\times\) 14)

= 528/43.988

= 12cm

The length of its minor axis (b) = 12cm

The volume of a cone (s) of height 6cm and base radius 5cm is

157cm^{3}

155cm^{3}

175cm^{3}

145cm^{3}

V = ^{1}/_{3 }πr^{2} h

Volume of a cone (Vc) = ^{1}/_{3 } \(\times\) 3.14 \(\times\) (5)^{2} \(\times\) 6

= 156.9cm^{3}

157cm^{3}

The probability of an outcome A is 1/6 . The probability of the B outcome is 1/4 . If the probability of A or B or both is 1/12 . What is the probability of both outcomes A and B?

1/2

1/3

2/5

3/4

Prob.(A outcome) = \(\frac{1}{6}\), Prob.(B outcome) = ¼

P(A ∩ B) = 1/12, Prob.(A \(\cup\) B)

Prob.(A \(\cup\) B) = Prob.(A) + Prob.(B) \(\cup\) Prob.(A ∩ B)

1/6 + 1/4 + 1/12

= (2 + 3 ? 1) ÷ 12

= 4/12

= 1/3

Prob.(A \(\cup\) B) = 1/3

Z = 2

_{n}= 2

_{4}

= 16

Calculate the value of x and y if 27^{x} ÷ 81^{x+2y} = 9 ,x + 4y = 0

x = 1, y = 1/2

x = 2, y = 1/2

x = 0, y = 1

x = 2, y = 1

\(27^x ÷ 81^{(x + 2y)} = 9 \)

\(27^x = 9 \times 81^{(x+2y)} \)

\({(3^{3} )}^x =32 \times 3^{4(x + 2y)} \)

\( =3^{(2 + 4x + 8y)}\)

\(3^{3x} = 3^{ (2 + 4x + 8y)}\)

3x = 2 + 4x + 8y

3x - 4x - 8y = 2 ----- (1)

x + 4y = 0 ------- (2)

- 4y = 2

y = (- 2) ÷ 4 = - ½

y = - ½

Substitute the value of y into equation (2)

i.e x + 4y = 0

x + 4( - 1/2) = 0

x - 2 = 0

x = 2

- x = 2,y = - ½

Method II

\( 27^x ÷ 31^{(x + 2y) }= 9\)

3^{3x} \(\times\) \(3^{( - 4x - 8y)} = 32\)

\( 3^{(3x - 8y)} = 32 \)

- x - 8y=2 ---- (1)

x + 4y = 0 ----- (2)

- 4 = 2

y= 2/4 = ½

y = ½

Substitute the value of y into equation 2

x + 4y=0

x + 4 (- 1) ÷ 2) = 0

x - 2 = 0

x = 2

x = 2, y = ½

Simplify \(\sqrt{30} \times \sqrt{40} \)

\(10 \sqrt{3}\)

\(5 \sqrt{3}\)

\(20 \sqrt{3}\)

\(15 \sqrt{3}\)

\(\sqrt{30} \times \sqrt{40}\)

\(\sqrt{30} \times \sqrt{40}\)

\(\sqrt{3 \times 10 \times 4 \times 10}\)

\(\sqrt{400 \times 3}\)

\(\sqrt{400 \times 3}\)

\(20 \times \sqrt{3}\)

\(20 \sqrt{3}\)

If ? (2x + 2) - -x =1 ,find x.

x = 1 twice

x = 0 or 1

x = 3 or 1

x = 2 twice

-(2x + 2 ) - -x = 1

-(2x + 2) = 1 + -x

Square both sides

-(2x + 2)^{2} = (1 + -x)^{2}

((2x + 2)^{2})^{½} = 1 + -x + -x+ x

2x + 2 = 1 + 2-x +x

Collect the like term together

2x - x + 2 - 1 = 2 -x

x + 1 = 2-x

Square both sides

(x + 1)^{2} = (2-x)^{2}

x^{2} + 2x + 1 = 4x

x^{2} + 2x - 4x + 1 = 0

(x^{2}- x)(x + 1)= 0

x(x - 1) -1(x - 1)

(x - 1)(x -1)

Either (x - 1) = 0 or (x - 1) = 0

x = 1 or 1

x = 1 twice

Then, the other is (5 ? x) , since their sum is 5 and their product is 14

x(5 ? x) = 14

5x ? x

^{2}= 14

x

^{2}? 5x ? 14 = 0

(x

^{2}? 7x) + (2x ? 14) = 0

x(x ? 7) + 2(x ? 7) = 0

(x ? 7)(x + 2) = 0

Either (x ? 7) = 0 or (x + 2) = 0

x = 7 or x = ? 2

The two numbers are 7 and ? 2

Solve the equation

\( 5^{(x - 2)} = (1 ÷ 125)^{(x +3)}\)

3/-2

-7/4

4/6

7/4

\( 5^{(x - 2)} = (125^{-1} )^{(x + 3)}\\

5^{(x - 2)} = (5- 3)^{x + 3}\\

5^{(x - 2)} = 5^{- 3x - 9}\\

x + 3x = - 9 + 2\\

4x = - 7\\

x = \frac{-7}{4} \)

How many sides has a regular polygon whose interior angles are 120°each

4

5

6

3

Each exterior angle = 180 - 120=60

Exterior angle =360/n

60 = 360/n

n =360/6

= 6

= N20.00

=

204

x 37

-----

7548 = 7.548

= 7.55(2dp)

A public car dealer marked up the cost of a car at 30% in an attempt to make 20% gross profit. Due to the value of dollar, he now placed 20% discount on the car. What profit or loss will he make?

3%

2%

4%

1%

Let assume the cost price is 100%

Marked up price + cost price = 20 + 100 = 120%

Discount at 20% = 20/100 \(\times\) 120 % of cost price

Selling price = cost price - gain

= (120 - 24)% of cost price

= 96% of cost price

Loss = (100- 96)% of cost price

= 4% of cost price

- He will wake 4% loss

Rationalize 5 ÷ (2 - \(\sqrt{3}\))

3(2 + \(\sqrt{5}\))

2(3 + \(\sqrt{5}\))

5(2 + \(\sqrt{3}\))

\(3\sqrt{3}\) + 1

5÷ (2 - \(\sqrt{3}\))

Using conjugate surds

[5 ÷ (2 - \(\sqrt{3}\))]\(\times\)(2 + \(\sqrt{3}\)) ÷ (2 + \(\sqrt{3}\))]

[5(2+\(\sqrt{3}\))÷((2- \(\sqrt{3}\)))2]

[5(2 +\(\sqrt{3}\)) ÷ (2)2- (\(\sqrt{3}\))2]

[5(2 +\(\sqrt{3}\)) ÷(4- 3)]

= 5(2 + \(\sqrt{3}\))

Factorize \( a^2 - b^2 - 4a + 4 \)

(a + b)(a - b)

(a - 2 + b)

(a + 1)(a - 2 + b)

(a + b) 2

The trinomial = \( a^2 - 4a + 4 \\

a^2 - b^2- 4a + 4 = (a^2 - 4a + 4) - b^2 \\

(a^2 - 2a - 2a + 4) - b^2 \\

[a(a - 2)- 2(a - 2)] - b^2 \\

(a - 2)2 - b^2 \\

(a - 2 + b)(a - 2 - b )\)

^{o}d" x d" 90

^{o}, Find the value of 2sinx.

thus m

^{2}= 2

^{2}+ 3

^{2}

= 4 + 9 = 13

m = \(\sqrt{13}\)

Hence, 2sin x = 2 x \(\frac{2}{m}\)

2 x\(\frac{2}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)

= \(\frac{4\sqrt{13}}{13}\)